$I=\int_{0}^{a} \frac{d x}{1+f(x)}$
$=\int_{0}^{a} \frac{1}{1+f(a-x)} d x$
$\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$=\int_{0}^{a} \frac{d x}{1+\frac{1}{f(x)}} \quad[\because f(x) f(a-x)=1]$
$I=\int_{0}^{a} \frac{f(x)}{f(x)+1} d x$
On adding Eqs. (i) and (ii), we get $2 I=\int_{0}^{a} \frac{f(x)+1}{f(x)+1} d x$
$\Rightarrow \quad 2 I=\int_{0}^{a} 1 d x$
$\begin{array}{ll}\Rightarrow & 2 t=[x]_{0}^{a} \\ \Rightarrow & 2 I=a \\ \Rightarrow & I=\frac{a}{2}\end{array}$
72. $(b,$ d) We have, $x=1-2 x$
$\Rightarrow y=\frac{1-2 x}{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-2 x-(1-2 x) \cdot 1}{x^{2}}$
$\quad=\frac{-2 x-1+2 x}{x^{2}}=\frac{-1}{x^{2}} < 0$
since, $a x+b y+c=0$ is tangent to the curve $x=1-2 x$
$\therefore \quad \frac{-a}{b} < 0 \Rightarrow \frac{a}{b}>0$
$\Rightarrow$ Either $a>a b>0$ or $a < 0, b < 0$